Download Multiplying Special Cases

Lesson 8-4 Warm-Up
ALGEBRA 1
“Multiplying Special Cases”
(8-4)
What are
“squares of a
binomial”?
Square of a Binomials: a binomial squared
Example: (a + b)2
How do you
square a
binomial?
Method 1: To square a binomial, multiply the binomial by itself. Then, use an
area model or the FOILing method.
Example: Simplify (a + b)2.
Method 2: RULE: To square a binomial, use the following formulas (the square
of the first term plus twice the product of the first and second term plus the
square of the last term):
(a + b)2 = a2 + 2ab + b2
ALGEBRA 1
(a - b)2 = a2 - 2ab + b2
Multiplying Special Cases
LESSON 8-4
Additional Examples
a. Find (y + 11)2.
(y + 11)2 = (y + 11)(y + 11)
= (y)2 + (2)y(11) + (11)2
= y2 + 22y + 121
Break into a multiplication problem
Substitute a with y and b with 11 in the rule
a2 + 2ab + b2.
Simplify.
b. Find (3w – 6)2.
= (3w)2 + 2(3w)(-6) + (-6)2 Substitute a with 3w and b with -6 in the
rule a2 + 2ab + b2.
= 9w2 – 36w + 36
Simplify.
ALGEBRA 1
Multiplying Special Cases
LESSON 8-4
Additional Examples
Among guinea pigs, the black fur gene (B) is dominant
and the white fur gene (W) is recessive. This means that a guinea
pig with at least one dominant gene (BB or BW) will have black fur.
A guinea pig with two recessive genes (WW) will have white fur.
The square below models the possible combinations of color genes that
parents who carry both genes can pass on to their offspring. Since WW
is 1 of the outcomes, the probability that a guinea pig has white fur is 1 .
4
4
B
B
W
BB BW
W BW WW
ALGEBRA 1
Multiplying Special Cases
LESSON 8-4
Additional Examples
(continued)
You can also model the probabilities found in the Punnett square with the
expression ( 1 B + 1 W)2. Show that this product gives the same result.
2
2
(12 B + 12 W)2 = ( 12 B)2 – 2(12 B)(12 W) + (12 W)2
1
1
1
= 4 B2 + 2 BW + 4 W 2
Square the binomial. Use
1
1
B
for
a
and
2 W for b
2
in (a + b)2 = a2 + 2ab + b2
Simplify.
The expressions 1 B2 and 1 W 2 indicate the probability that offspring
4
4
will have either two dominant genes or two recessive genes is 1 . The
4
1
1
expression BW indicates that there is chance that the offspring
2
2
will inherit both genes. These are the same probabilities shown in the
model.
ALGEBRA 1
Multiplying Special Cases
LESSON 8-4
Additional Examples
a. Find 812 using mental math.
812 = (80 + 1)2
= (80)2 + 2(80)(1) + (1)2
Substitute 80 for a and 1 for b in the rule
(a + b)2 = a2 + 2ab + b2.
= 6400 + 160 + 1 = 6561
Simplify.
b. Find 592 using mental math.
592 = (60 – 1)2
= 602 + 2(60)(-1) + (-1)2
Substitute 80 for a and 1 for b in the rule
(a + b)2 = a2 + 2ab + b2.
= 3600 – 120 + 1 = 3481
Simplify.
ALGEBRA 1
“Special Cases”
(8-4)
What is the
“difference of
squares”?
Difference of Squares: the product of the sum and difference of the same two
terms
Example: (a + b)(a - b)
How do you find Rule: To find the difference of squares, use an area model or the FOILing
the “difference of method. The FOILing method is used below.
squares”?
(a + b)(a - b) = a(a) + a(-b) + b(a) + b(-b) = a2 – ab + ab - b2
= a2 - b2
Notice that the products of the Outside and Inside of FOIL cancel each other out
(in other words, equal zero). So, the product of the sum and difference of the
same two terms is the difference of their squares.
ALGEBRA 1
Multiplying Special Cases
LESSON 8-4
Additional Examples
Find (p4 – 8)(p4 + 8).
(p4 – 8)(p4 + 8) = (p4)2 – (8)2
= p8 – 64
Use (a + b)(a - b) = a2 - b2 where a = p4
and b = 8 .
Simplify.
ALGEBRA 1
Multiplying Special Cases
LESSON 8-4
Additional Examples
Find 43 • 37.
43 • 37 = (40 + 3)(40 – 3)
Express each factor using 40 and 3. This
models (a + b)(a - b) = a2 - b2 where a = 40
and b = 3 .
= (40)2 – (3)2
Use (a + b)(a - b) = a2 - b2
= 1600 – 9 = 1591
Simplify.
ALGEBRA 1
Multiplying Special Cases
LESSON 8-4
Lesson Quiz
Find each square.
1. (y + 9)2
y2 + 18y + 81
2. (2h – 7)2
4h2 – 28h + 49
3. 412
4. 292
1681
5. Find (p3 – 7)(p3 + 7).
p6 – 49
841
6. Find 32 • 28.
896
ALGEBRA 1